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The athletes run a total distance of 300 m.
Total Distance vs Total Displacement Example:
Defined as: The rate of change of velocity.
It is a vector quantity measured in m s-2.
Where v = final velocity, u = initial velocity.
A professor walks path ABCDA around her garden.
Calculate Distance: Sum of all sides: 15 + 9 + 15 + 9 = 48 km.
Calculate Displacement: She travels back to A. Distance from original position is 0. Displacement = 0 km.
Constant Velocity
Increasing Velocity
Const Velocity (a=0)
Const Acceleration
Increasing Accel.
To find displacement: Split area into shapes (triangles/rectangles).
Area of triangle = 12 x base x height
Area of rectangle = base x height
Variables (SUVAT):
s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time
From the gradient of a velocity-time graph:
Rearrange: at = v - u → v = u + at
Displacement is the area under v-t graph.
Average velocity is halfway between u and v.
Area under graph = Rectangle (ut) + Triangle (12(v-u)t).
Since (v-u) = at:
Substitute t = (v-u)/a into equation 2.
Train speed 50 m s-1. Must pass marker 2 at 10 m s-1. Time limit 20 s.
Knowns: u = 50, v = 10, t = 20, s = ?
Equation: s = u + v2 t
Calculation: s = 50 + 102 x 20 = 600 m
Calculate 'g' by measuring time for a ball-bearing to fall a distance.
Use equation: s = ut + 12at2
Rearranged for straight line graph: 2h/t = gt + 2u
Vertical and horizontal components are independent.
Horizontal (→)
a = 0
ux = u cos θ
Range R = (u cos θ)t
Vertical (↑)
a = -g
uy = u sin θ
Use SUVAT
Time of Flight: t = 2u sin θg
Max Height: H = (u sin θ)22g
Range: R = u2 sin 2θg
Takes off horizontally 1.25m high. Lands 10m away. What was speed?
Step 1: Vertical (Find Time)
s = 1.25, u = 0, a = 9.81
s = 0.5gt2 → t = 2s/g = 0.5 s
Step 2: Horizontal (Find Speed)
s = 10, t = 0.5, a = 0
u = s/t = 10/0.5 = 20 m s-1